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I'm reading chapter 11 of The Blackwell Guide to the Philosophy of Language by Frank Jackson, and once he touched upon Lewis' triviality results he writes:

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However, this definitely doesn't look that simple to me. Could you break down this proof line by line and show why it holds?

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The argument can be outlined as follows (I presume it is not hard to supply philosophical and mathematical details):

Ex hypothesi, P(A → B) = P(B | A) = P(A & B) / P(A).

By the rule of total probability, we have

(*) P(A → B) = P(A → B | B) P(B) + P(A → B | ~B) P(~B)

where the propositions B and ~B are complementary (instead, one may take any relevant set of mutually exclusive and collectively exhaustive propositions).

Consider multiple conditioning (i.e., “conditionalising”): Suppose we have P(A | B) and want to put a further condition C; that is, we seek the probability of a proposition A given B, given C. In ill-formed notation (just to illustrate the idea!), what we want is P(A | B | C). Properly, this is P(A | B & C) which is equal to P(A & B & C) / P(B & C) – you can easily check these by the conditional probability formula.

We apply multiple conditioning to P(A → B) with B and ~B:

P(A → B | B) = P( B | A & B)

P(A → B | ~B) = P( B | A & ~B)

Substituting into (*), we get

P(A → B) = P(B | A & B) P(B) + P(B | A & ~B) P(~B)

Observe that

P(B | A & B) = P(B & A & B) / P(A & B) = P(A & B) / P(A & B) = 1

P(B | A & ~B) = P(B & A & ~B) / P(A & ~B) = P(⊥) / P(A & ~B) = 0

So, we get P(A → B) = P(B | A) = P(B)

Therefore, the equation P(A & B) = P(A)P(B) holds true and propositions A and B are independent, while, according to the argument, the probability of B has to be dependent on the probability of A in general.

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